Given `y=2cosx+2sinx+ln(cosx)+x` , find the intervals of increase and decrease on `[0,2pi]` .

(1) Take the first derivative and set equal to zero to find the critical points.

`y'=-2sinx + 2cosx-(sinx)/(cosx)+1=0` Multiply through by cosx

`-2sinxcosx+2cos^2x-sinx+cosx=0` Rearrange terms

`2cos^2x+(-2sinx+1)cosx-sinx=0` This is a quadratic in cosx; Use the quadratic formula:

`cosx=((2sinx-1)+- sqrt((-2sinx+1)^2-4(2)(-sinx)))/4`

`cosx=((2sinx-1)+-sqrt(4sin^2x-4sinx+1+8sinx))/4`

`cosx=((2sinx-1)+-sqrt((2sinx+1)^2))/4`

`cosx=((2sinx-1)+(2sinx+1))/4` or `cosx=((2sinx-1)-(2sinx+1))/4`

`cosx=sinx` or `cosx=-1/2`

On `[0,2pi]` this gives solutions of `x=pi/4,(5pi)/4,(2pi)/3,(4pi)/3` .

However, the function is not defined on `[pi/2,(3pi)/2]` since cos is nonpositive on this interval, and ln is undefined for nonpositive inputs.

Thus we need only check ` ` `[0,pi/2)` and `((3pi)/2,2pi]` .The only critical value on these intervals is `pi/4` . A quick check shows `y'>0` for `0<x<pi/4` and `y'<0` for `pi/4<x<pi/2` and `y'>0` for `x>(3pi)/2` .

Thus the function is increasing on `[0,pi/4)` , decreasing on `(pi/4,pi/2)` , undefined on `[pi/2,(3pi)/2]` , and increasing on `(3pi)/2,2pi]` .